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Hello, I would like to know if the following result is true:

Let $A,B$ be two embedded circles in $S^2$ which do not intersect and let $C$ be the $\textit{closed}$ region bounded by $A$ and $B$ (this region is well-defined by the Jordan Curve Theorem). Then $C$ is homeomorphic to $S^1\times [0,1]$.

If so, is there an easy proof, or does this result have a name?

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This is the annulus theorem, and is indeed true for circles in $S^2$ without need for further hypotheses (which are needed in higher dimensions).


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Allen: thanks for fixing the broken link! – tweetie-bird Jun 25 '12 at 1:15

A more general result is true. Let F and G be two disjoint connected compact subsets of the sphere. Let D be the component of the complement whose boundary intersects both. Then D is conformally equivalent to a standard ring (or a punctured disc, or the punctured plane). This theorem can be easily derived from the Riemann mapping theorem in the simply connected case. See, for example the book by Goluzin, Geometric theory of functions of complex variable.

In your case, then F and G are Jordan curves, one also needs to extend this homeomorphism to the boundary. This can be done with the same technique as for the Riemann case. The boundary extension is a local question.

Of course, a purely topological proof, not using complex variables must also exist.

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If you are willing to quote the Schoenflies theorem and the classification of surfaces, a quick proof of this result is a standard exercise. First, using algebraic topology (as in the proof of the Jordan curve theorem itself), one shows that $S^1 - (A \cup B)$ is a union of three components, whose closures are three compact subsets $S_A$, $S_{AB}$, $S_B$ with frontiers $A$, $A \cup B$, $B$ respectively. Applying the Schoenflies theorem, each of $S_A$, $S_{AB}$, $S_B$ is a compact surface with boundary, and in fact $S_A$, $S_B$ are homeomorphic to a disc. Next one has $\chi(S_A) + \chi(S_{AB}) + \chi(S_B) = \chi(S^2)$, which is a standard result about gluing together a bunch of compact surfaces-with-boundary by pairwise identifying boundary circles. So that equation becomes $1 + \chi(S_{AB}) + 1 = 2 \implies \chi(S_{AB})=0$. Using the classification of surfaces, and ruling out all compact surfaces of Euler characteristic zero with other than two boundary components, $S_{AB}$ is an annulus.

The same technique carried out in general allows one to take any compact surface $S$ with boundary (specified by its Euler characteristic, orientability, and number of boundary components), any integer $n$, and any subset $\mathcal{C} \subset S$ consisting of $n$ disjointly embedded circles, and enumerate all of the finitely many possibilities for the topological types of the components of $S-\mathcal{C}$.

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About the first step of your proof: actually, by Schoenflies applied to $A$, the component $L$ of $S^2-A$ containing $B$ is an open disc, and applying now the Schoenflies theorem to $B$ in the one-point compactification of $L$ shows that the component of $S^2-(A\cup B)$ touching both $A$ and $B$ is homeomorphic to an open cylinder. – YCor Aug 4 '12 at 23:11

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